Question: Determine the number of angles between 0 and $2 \pi,$ other than integer multiples of $\frac{\pi}{2},$ such that $\sin \theta,$ $\cos \theta$, and $\tan \theta$ form a geometric sequence in some order.
Solution: We divide into cases.

Case 1: $\sin \theta \tan \theta = \cos^2 \theta.$

The equation becomes $\sin^2 \theta = \cos^3 \theta,$ which we can write as $1 - \cos^2 \theta = \cos^3 \theta.$  Letting $x = \cos \theta,$ we get
\[x^3 + x^2 - 1 = 0.\]Let $f(x) = x^3 + x^2 - 1.$  Clearly $x = -1$ is not a root.  If $-1 < x \le 0,$ then $x^2 + x^3 \le x^2 < 1$, so
\[f(x) = x^3 + x^2 - 1 < 0.\]The function $f(x)$ is increasing for $0 \le x \le 1.$  Also, $f(0) = -1$ and $f(1) = 1,$ so $f(x)$ has exactly one root in the interval $[0,1].$  Then the equation $\cos \theta = x$ has two solutions for $0 \le \theta \le 2 \pi.$

Case 2: $\sin \theta \cos \theta = \tan^2 \theta.$

The equation becomes $\cos^3 \theta = \sin \theta.$  In the interval $0 \le \theta \le \frac{\pi}{2},$ $\sin \theta$ increases from 0 to 1 while $\cos^3 \theta$ decreases from 1 to 0, so there is one solution in this interval.  Similarly, in the interval $\pi \le \theta \le \frac{3 \pi}{2},$ $\sin \theta$ decreases from 0 to $-1$ while $\cos^3 \theta$ increases from $-1$ to $0,$ so there is one solution in this interval.

On the intervals $\frac{\pi}{2} < \theta < \pi$ and $\frac{3 \pi}{2}  < \theta < 2 \pi,$ one of $\sin \theta$ and $\cos^3 \theta$ is positive while the other is negative, so there are no additional solutions.

Case 3: $\cos \theta \tan \theta = \sin^2 \theta.$

The equation becomes $\sin \theta^2 = \sin \theta$, so $\sin \theta$ is 0 or 1.  The only solutions are integer multiples of $\frac{\pi}{2},$ so there are no solutions in this case.

Therefore, there are a total of $\boxed{4}$ solutions.